[英]using shared_ptr to std::vector in range-based for loop
I wrote a c++ function that assembles some data then then returns a std::shared_ptr
to a newly allocated std::vector
containing the data. 我编写了一个c ++函数,它汇总了一些数据,然后将std::shared_ptr
返回给包含数据的新分配的std::vector
。 Something analogous to this: 类似于此的东西:
std::shared_ptr<std::vector<int>> shared_ptr_to_std_vector_of_ints()
{
auto v = std::make_shared<std::vector<int>>();
for (int i = 0; i < 3; i++) v->push_back(i);
return v;
}
I tried to iterate over the contents of the vector using a range-based for loop, but it acted as if the vector was empty. 我尝试使用基于范围的for循环迭代向量的内容,但它的作用就好像向量是空的。 After fiddling around, I found I could get it to behave as I expected by assigning the value returned from the function to a local variable, then referencing that in the loop: 摆弄后,我发现通过将函数返回的值赋给局部变量,然后在循环中引用它,我可以让它按照我的预期运行:
// Executes loop zero times:
std::cout << "First loop:" << std::endl;
for (int i : *shared_ptr_to_std_vector_of_ints()) std::cout << i << std::endl;
// Prints three lines, as expected
std::cout << "Second loop:" << std::endl;
auto temp = shared_ptr_to_std_vector_of_ints();
for (int i : *temp) std::cout << i << std::endl;
That snipped prints this: 剪断打印这个:
First loop:
Second loop:
1
2
3
Why does the first version not work? 为什么第一个版本不起作用?
I'm using Xcode on macOS Sierra 10.12.6. 我在macOS Sierra 10.12.6上使用Xcode。 I believe it is using LLVM 9.0 to compile the c++ code. 我相信它正在使用LLVM 9.0来编译c ++代码。
Note that shared_ptr_to_std_vector_of_ints
returns by-value, so what it returns is a temporary. 请注意, shared_ptr_to_std_vector_of_ints
按值返回,因此它返回的是临时值。
The range-based for loop is equivalent to 基于范围的for循环等效于
{
init-statement
auto && __range = range_expression ;
auto __begin = begin_expr ;
auto __end = end_expr ;
for ( ; __begin != __end; ++__begin) {
range_declaration = *__begin;
loop_statement
}
}
The part auto && __range = range_expression ;
零件auto && __range = range_expression ;
, for your example it will be auto && __range = *shared_ptr_to_std_vector_of_ints() ;
,对于你的例子,它将是auto && __range = *shared_ptr_to_std_vector_of_ints() ;
. 。 shared_ptr_to_std_vector_of_ints
returns a temporary std::shared_ptr<std::vector<int>>
, then dereference on it to get the std::vector<int>
, then bind it to the rvalue-reference __range
. shared_ptr_to_std_vector_of_ints
返回一个临时的std::shared_ptr<std::vector<int>>
,然后取消引用它以获取std::vector<int>
,然后将其绑定到rvalue-reference __range
。 The temporary std::shared_ptr<std::vector<int>>
will be destroyed after the full expression, and use_count
decreases to 0 so the std::vector<int>
being managed is destroyed too. 在完整表达式之后,临时std::shared_ptr<std::vector<int>>
将被销毁,并且use_count
减少到0,因此被管理的std::vector<int>
也被销毁。 Then __range
becomes dangled reference. 然后__range
变成__range
参考。 After that eg auto __begin = begin_expr ;
之后例如auto __begin = begin_expr ;
would try to get the iterator from __range
, which leads to UB. 会尝试从__range
获取迭代器,这会导致UB。
(emphasis mine) (强调我的)
If range_expression returns a temporary, its lifetime is extended until the end of the loop, as indicated by binding to the rvalue reference
__range
, but beware that the lifetime of any temporary within range_expression is not extended. 如果range_expression返回临时值,则其生命周期将延长到循环结束,如绑定到右值引用__range
,但要注意__range
中任何临时值的生命周期都不会延长。
As your 2nd version showed, the issue could be solved via using a named variable instead; 正如您的第二个版本所示,问题可以通过使用命名变量来解决; or you can also use init-statement (from C++20): 或者您也可以使用init-statement(来自C ++ 20):
for (auto temp = shared_ptr_to_std_vector_of_ints(); int i : *temp) std::cout << i << std::endl;
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